June 2014 (R) MS - C2 Edexcel
Transcript of June 2014 (R) MS - C2 Edexcel
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Edexcel and BTEC Qualifications
Edexcel and BTEC qualifications are awarded by Pearson, the UK’s largest awarding body. We
provide a wide range of qualifications including academic, vocational, occupational and specific
programmes for employers. For further information visit our qualifications websites atwww.edexcel.com or www.btec.co.uk. Alternatively, you can get in touch with us using the
details on our contact us page at www.edexcel.com/contactus.
Pearson: helping people progress, everywhere
Pearson aspires to be the world’s leading learning company. Our aim is to help everyoneprogress in their lives through education. We believe in every kind of learning, for all kinds of
people, wherever they are in the world. We’ve been involved in education for over 150 years,
and by working across 70 countries, in 100 languages, we have built an international reputation
for our commitment to high standards and raising achievement through innovation in education.
Find out more about how we can help you and your students at: www.pearson.com/uk
Summer 2014
Publications Code UA038458
All the material in this publication is copyright© Pearson Education Ltd 2014
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General Marking Guidance
• All candidates must receive the same treatment. Examiners
must mark the first candidate in exactly the same way as theymark the last.
• Mark schemes should be applied positively. Candidates mustbe rewarded for what they have shown they can do ratherthan penalised for omissions.
• Examiners should mark according to the mark scheme notaccording to their perception of where the grade boundariesmay lie.
• There is no ceiling on achievement. All marks on the markscheme should be used appropriately.
• All the marks on the mark scheme are designed to beawarded. Examiners should always award full marks ifdeserved, i.e. if the answer matches the mark scheme.Examiners should also be prepared to award zero marks if thecandidate’s response is not worthy of credit according to themark scheme.
• Where some judgement is required, mark schemes will providethe principles by which marks will be awarded andexemplification may be limited.
•
Crossed out work should be marked UNLESS the candidate hasreplaced it with an alternative response.
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PEARSON EDEXCEL GCE MATHEMATICS
General Instructions for Marking
1. The total number of marks for the paper is 75
2. The Edexcel Mathematics mark schemes use the following types of marks:
• M marks: Method marks are awarded for ‘knowing a method and attempting to
apply it’, unless otherwise indicated.
• A marks: Accuracy marks can only be awarded if the relevant method (M) marks
have been earned.
• B marks are unconditional accuracy marks (independent of M marks)
• Marks should not be subdivided.
3.
Abbreviations
These are some of the traditional marking abbreviations that will appear in the markschemes.
• bod – benefit of doubt
• ft – follow through
• the symbol will be used for correct ft
• cao – correct answer only
•
cso - correct solution only. There must be no errors in this part of the question toobtain this mark
• isw – ignore subsequent working
• awrt – answers which round to
• SC: special case
• oe – or equivalent (and appropriate)
• d… or dep – dependent
• indep – independent
•
dp decimal places• sf significant figures
• The answer is printed on the paper or AG - answer given
• or d… The second mark is dependent on gaining the first mark
• aliter – alternative method
• aef – any equivalent form
4. All A marks are ‘correct answer only’ (cao.), unless shown, for example, as A1 ft to
indicate that previous wrong working is to be followed through. After a misreadhowever, the subsequent A marks affected are treated as A ft, but manifestly absurdanswers should never be awarded A marks.
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5. For misreading which does not alter the character of a question or materially simplifyit, deduct two from any A or B marks gained, in that part of the question affected.
6. 7. If a candidate makes more than one attempt at any question:
• If all but one attempt is crossed out, mark the attempt which is NOT crossed
out.• If either all attempts are crossed out or none are crossed out, mark all theattempts and score the highest single attempt.
8. Ignore wrong working or incorrect statements following a correct answer.
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General Principles for Core Mathematics Marking (But note that specific mark schemes may sometimes override these general principles).
Method mark for solving 3 term quadratic:
1. Factorisation
c pqq x p xcbx x =++=++ where),)(()( 2 , leading to x = …
amnc pqqnx pmxcbxax ==++=++ and where),)(()( 2 , leading to x = …
2. Formula
Attempt to use the correct formula (with values for a, b and c).
3. Completing the square
Solving 02 =++ cbx x : 0,02
2
≠=±±
± qcq
b x , leading to x = …
Method marks for differentiation and integration:
1. Differentiation
Power of at least one term decreased by 1. ( 1−→ nn x x )
2. Integration
Power of at least one term increased by 1. ( 1+→ nn x x )
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Use of a formula
Where a method involves using a formula that has been learnt, the advice given in recent
examiners’ reports is that the formula should be quoted first.
Normal marking procedure is as follows:
Method mark for quoting a correct formula and attempting to use it, even if there are smallerrors in the substitution of values.
Where the formula is not quoted, the method mark can be gained by implication from
correct working with values, but may be lost if there is any mistake in the working.
Exact answers
Examiners’ reports have emphasised that where, for example, an exact answer is asked for,
or working with surds is clearly required, marks will normally be lost if the candidate resorts
to using rounded decimals.
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Question
NumberScheme Marks
1.
83
12
x +
1 12 x+ Both terms correct as printed (allow 112 x
but not 81 )
B1
2 3
2 3
8 8
2 3
8(7) 3 8(7)(6) 3... ...
2! 2 3! 2
3 3... C C ...
2 2
x x
x x
+ + +
+ + +
2 38(7) 8(7)(6)
... or ...2! 3! x x
× × × ×
or
( ) ( )8 2 8 32 3C ... or C ... x x× × × ×
M1: For either the x2 term or the 3 x term.
Requires correct binomial coefficient in
any form with the correct power of x, but
the other part of the coefficient (perhaps
including powers of 2 and/or 3 or signs)
may be wrong or missing.
M1
Special Case: Allow this M1 only for an attempt at a descending expansion
provided the equivalent conditions are met for any term other than the first
e.g.
( ) ( )7 6
2
7 6
8 8
1 2
3 8(7) 3... 8 1 1 ...
2 2! 2
3 3... C C ...
2 2
x x
x x
+ + +
+ + +
2 3... 63 189 ... x x+ + + A1: Either 263 x or 3189 x
A1A1A1: Both 263 x and 3189 x
Terms may be listed but must be positive
[4]
Total 4
Note it is common not to square the 2 in the denominator of3
2
x
and this gives
2 31 12 126 756 x x x+ + + . This could score B1M1A0A0.
Note
2 3
8 4 8 3
2 3
3 3... C 1 C 1 ...
2 2
x x + + +
would score M0 unless a correct method
was implied by later work
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Question
NumberScheme Marks
S 6a∞ =
2. (a)
61
aa
r =
− Either 6
1
aa
r =
− or
6
1
aa
r =
− or
61
1 r =
− M1
{ } 5
1 6(1 ) *6
r r ⇒ = − ⇒ = cso A1*
Allow verification e.g.5 16 6
6 6 6 6 61 1
a a aa a a a ar
= ⇒ = ⇒ = ⇒ =− −
[2]
(b){ }
3
3
4
5T 62.5 62.5
6ar a
= = ⇒ =
35
62.56
a
=
(Correct statement using
the 4th term. Do not accept4
562.5
6a
=
)
M1
108a⇒ = 108 A1
[2]
(c)
{ }their
S 6(their ) or 6485
16
aa∞ = =
−
Correct method to find S∞ M1
{ } ( )
{ }305
6
30 56
108 1 ( )S 645.2701573...
1
−= =
−
M1:
( )30
30
5their 1
6S
51
6
a
− =
−
(Condone invisible brackets around
5/6)M1 A1ft
A1ft: Correct follow through expression(follow through their a). Do not
condone invisible brackets around 5/6
unless their evaluation or final answer
implies they were intended.
{ }30S S 2.72984...∞ − = awrt 2.73 A1
[4]
Total 8
(c)
Alternative:
Difference =
30
30
5108
6 2.72984...51
16
ar r
= =− −
M1M1: For an attempt to apply30
1
ar
r −.
A1ft:( ) 30
1
their a r
r
×
− with their ft a.
A1: awrt 2.73
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Question Number
Scheme Marks
3. (a) 7 and 15
Both 7 and 15 .
Allow awrt 2.65 and 3.87B1
[1]
(b) ( ) }{1Area ( ) 2; 3 2 7 11 15 192
R ≈ × × + + + +
Outside brackets 12
2× or 1
(may be implied)B1;
For structure of { }.............
M1
Note decimal values are
( )}{ }{1 12; 3 19 2 7 11 15 2; 6.0909.. 19.6707...2 2
× × + + + + = × × +
M1 requires the correct structure for the y values. It needs to contain first y value plus
last y value and the second bracket to be multiplied by 2 and to be the summation of the
remaining y values in the table with no additional values. If the only mistake is a
copying error or is to omit one value from 2(…..) bracket this may be regarded as a slip
and the M mark can be allowed (nb: an extra repeated term, however, forfeits the M
mark). M0 if any values used are x values instead of y values.
Bracketing mistakes: e.g.
( ) ( )1 2 3 19 2 7 11 152
× × + + + +
( )1 2 3 19 2 7 11 152
× × + + + +
Both score B1 M1
Alternative:Separate trapezia may be used, and this can be marked equivalently.
1 1 1 1
2( 3 7) 2( 7 11) 2( 11 15) 2( 15 19)2 2 2 2
× + + × + + × + + × +
B1 for1
22
× , M1 for correct structure
1 25.76166865... 25.76166... 25.76= × = = (2dp) 25.76 A1 cao
[3]
(c) underestimate Accept ‘under’, ‘less than’ etc. B1
[1]
Total 5
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Question
NumberScheme Marks
4. (a)
3 2f ( ) 4 9 18 x x a x x= − + + −
f (2) 32 4 18 18 0
4 32 8
a
a a
= − + + − =
⇒ = ⇒ =
Attempts f(2) or f ( 2 )− M1
cso A1
[2]
(a)
Way 2
( )( )2f ( ) 2 x x px qx r = − + +
( ) ( )3 22 2 2 px q p x r q x r = + − + − −
9 0 also 4 2 8r q p a p= ⇒ = = − ∴ = − = Compares coefficients leading
to
-2 p = a
M1
8a = cso A1
(a)
Way 3
( ) ( )3 24 9 18 2 x a x x x− + + − ÷ −
( )24 8 2 7Q x a x a= − + − + − 4 32 R a= −
Attempt to divide ±f( x) by ( x –
2) to give a quotient at least of
the form 24 g( ) x a x± + and a
remainder that is a function of a
M1
4 32 0 8a a− = ⇒ = cso A1
(b)
2f ( ) ( 2)( 4 9) x x x= − − +
Attempts long division or other
method, to obtain2( 4 ), x ax b− ± ± 0,b ≠ even
with a remainder. Working
need not be seen as this could
be done “by inspection.”
M1
( 2)(3 2 )(3 2 )
orequivalente.g.
( 2)(2 3)(2 3)
or
( 2)(2 3)( 2 3)
x x x
x x x
x x x
= − − +
= − − − +
= − − − −
dM1: A valid attempt to
factorise their quadratic – see
General Principles. This isdependent on the previous
method mark being awarded,
but there must have been no
remainder.dM1A1
A1: cao – must have all 3
factors on the same line. Ignore
subsequent work (such as a
solution to a quadratic
equation.)
[3]
(c)
1 1 1 1f 4 8 9 18 12
2 8 4 2
= − + + − = −
Attempts ( )12f or ( )12
f −
M1A1ft Allow A1ft for the correct
numerical value oftheir
144
a−
[2]
(c)
Way 2
± ( ) ( )3 24 8 9 18 2 1 x x x x− + + − ÷ −
22 3 6Q x x= − + +
12 R = −
M1: Attempt long division to
give a remainder that is
independent of x
M1A1ft A1: Allow A1ft for the correct
numerical value oftheir
14.4
a−
Total 7
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Question
NumberScheme Marks
5(a)Length DEA 7(2.1) 14.7= =
M1: 7 2.1× onlyM1A1
A1: 14.7
[2]
(b)
Angle CBD 2.1π = −
May be seen on the diagram (allow awrt 1.0 and allow 180 –
120). Could score for sight ofAngle CBD = awrt 60 degrees.
M1
Both 7cos( 2.1)π − and 7sin( 2.1)π −
or
Both 7cos( 2.1)π − and ( )227 7 cos( 2.1)π − −
or
Both 7sin( 2.1)π − and ( )227 7 sin( 2.1)π − −
Or equivalents to these
A correct attempt to find BC and
BD. You can ignore how the
candidate assigns BC and CD.
7cos( 2.1)π − can be implied by
awrt 3.5 and 7sin( 2.1)π − can be
implied by awrt 6. Note if the sin
rule is used, do not allow mixing of
degrees and radians unless their
answer implies a correct
interpretation. Dependent on theprevious method mark.
dM1
Note that 2.1 radians is 120 degrees (to 3sf) which if used gives angle CBD as 60
degrees. If used this gives a correct perimeter of 31.3 and could score full
marks.
P 7 cos( 2.1) 7sin( 2.1) 7 14.7= π − + π − + +
their BC + their CD + 7 + their
DEA
Dependent on both previous
method marks
ddM1
31.2764...= Awrt 31.3 A1
[4] Total 6
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Question
NumberScheme Marks
6.
{ }4 3
3 21 3d
8 4 32 4
x x x x x c
+ = + +
∫ M1: 1n n x x +→ on either term
M1A1A1:4 3
.32 4
x x+ Any correct
simplified or un-simplified
form. (+ c not required)2
4 3
4
16 8 256 ( 64)
32 4 32 4 32 4
x x
−
− + = + − +
or
( ) ( ) ( )
( )0 24 44 3 3 4 3 3
4 0
4 2( 4) (2)0 added to 0
32 4 32 4 32 4 32 4
x x x x
−
− − + = − + + = + −
dM1
Substitutes limits of 2 and 4− into an “integrated function” and subtracts either
way round. Or substitutes limits of 0 and 4− and 2 and 0 into an “integrated
function” and subtracts either way round and adds the two results.
21
2=
21
2 or 10.5 A1
{ }At 4, 8 12 4 or at 2, 1 3 4 x y x y= − = − + = = = + =
Area of Rectangle 6 4 24= × =
or
Area of Rectangles 4 4 16= × = and 2 4 8× = M1
Evidence of ( ) 44 2 their y−− − × or ( ) 24 2 their y− − ×
or
Evidence of44 their y−× and 22 their y×
So, 21 27
area(R) 242 2
= − =
dddM1: Area rectangle –
integrated answer. Dependenton all previous method marks
and requires:
Rectangle > integration > 0 dddM1A1
A1:27
2 or 13.5
[7]
Total 7
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Alternative:
3 21 3"their4" d 8 4
x x x
± − + ∫
Line – curve. Condone missing
brackets and allow either way
round.
4th M1
= { }4 3
432 4
x x x c− − +
M1: 1n n x x +→ on either curve
term
1st M1,1st
A1ft
A1ft:4 3
" ."32 4 x x− − Any correct
simplified or un-simplified form
of their curve terms, follow
through sign errors. (+ c not
required)
[ ]2
4
16 8 256 ( 64)8 16
32 4 32 4−− = − − − − − −
2nd M1 Substitutes limits of 2
and 4− into an “integrated
curve” and subtracts either way
round.
2nd
M1, 3rd
M1
2nd A1
3rd M1 for ±("8" " 1 6"− − )
Substitutes limits into the ‘line
part’ and subtracts either way
round.
2nd A1 for correct ± (underlined
expression). Now needs to be
correct but allow ± the correct
expression.
27
2= A1:
27
2 or 13.5 3rd A1
If the final answer is -13.5 you can withhold the final A1
If -13.5 then “becomes” +13.5 allow the A1
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Question
NumberScheme Marks
7.(i)
sin21
(4 sin 2 1)
θ
θ =
−; 0 180θ °
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Question
NumberScheme Marks
5 8 y =
8. (i)
log5 log8 y = log5 log8 y = or 5log 8 y = M1
log81.2920...
log5 y
= =
awrt 1.29 A1
Allow correct answer only
[2]
( ) 12 22log 15 4 log x x+ − =
(ii)
( ) 1
2
2 2log 15 4 log x x+ − = Applies the power law of logarithms
seen at any point in their working M1
12
2
15log 4
x
x
+ =
Applies the subtraction or addition law
of logarithms at any point in their
working
M1
12
415
2
x
x
+
=
Obtains a correct expression with logs
removed and no errors M112
2
16 15 0
or e.g.
225 226
x x
x x
− + =
+ =
Correct three term quadratic in any
formA1
( )( )1 15 0 ... x x x− − = ⇒ = A valid attempt to factorise or solve
their three term quadratic to obtain
... x = or x = ... Dependent on all
previous method marks.
dddM1
{ }1,15 x =
1, 225 x = Both x = 1 and x = 225 (If both are
seen, ignore any other values of 0 x ≤
from an otherwise correct solution)
A1
[6]
Total 8
Alternative:
( )2 22log 15 8 log x x+ − =
( )2
2 2log 15 8 log x x+ − = Applies the power law of logarithms M1
2
2
( 15)
log 8
x
x
+
= Applies the subtraction law of
logarithms M1
28( 15) 2
x
x
+=
Obtains a correct expression with logs
removedM1
2 30 225 256 x x x+ + =
2 226 225 0 x x− + = Correct three term quadratic in any
formA1
( )( )1 225 0 ... x x x− − = ⇒ = A valid attempt to factorise or solve
their 3TQ to obtain ... x = Dependent
on all previous method marks.
dddM1
1, 225 x = Both x
= 1 and x
= 225 (If both areseen, ignore any other values of 0 x ≤
from an otherwise correct solution)
A1
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Question
NumberScheme Marks
9. (a)
{ }2
21sin60
2 2 2
x A xy x
π ° = + +
M1: An attempt to find 3 areas of the
form:2 2, and xy p x qxπ M1A1
A1: Correct expression for A (terms
must be added)
( )2 2
3 50 3 5050 2 38 4 8 4 8 x x x x x xy y y
x xπ π π = + + ⇒ = − − ⇒ = − + *
Correct proof with no errors seen
A1 *
[3]
(b)
{ } 2 22
xP x y
π = + +
Correct expression for P in terms of x
and y B1
( )50
2 2 2 32 8
x xP x
x
π π
= + + − + √
Substitutes the given expression for y
into an expression for P where P is at
least of the form x yα β +
M1
100 3 100 3
2 22 4 2 4 2
x x x
P x x P x x x x
π π π
= + + − − ⇒ = + + −
( )100
8 2 34
xP
xπ ⇒ = + + − Correct proof with no errors seen A1 *
[3]
(Note8 2 3
4
π + −= 1.919……)
2d 8 2 3100d 4
P x
x
π − + −= − +
M1: Either xµ µ → or2
100
x x
λ ±→
M1A1
(c) and
(d)
can be
marked
together
A1: Correct differentiation (need not be
simplified). Allow ( )2
100 awrt1 92 x−
− + ⋅
2 8 2 3100 0 ...4
x xπ − + −− + = ⇒ =
Their 0P′ = and attempt to solve as far as
x = …. (ignore poor manipulation)M1
4007.2180574...
8 2 3 x
π ⇒ = =
+ −
400
8 2 3π + − or awrt 7.2 and no other
values
A1
{ }7.218..., x = 27.708... (m)P⇒ = awrt 27.7 A1[5]
2
2 3
d 2000 Minimum
d
P
x x= > ⇒
M1: Finds P′′ ( 1n n x x −→ allow for
constant 0→ ) and considers sign
M1A1ftA1ft:
3
200
x(need not be simplified) and
> 0 and conclusion. Only follow through
on a correct P′′ and a single positivevalue of x found earlier.
[2]
Total 13
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Question
NumberScheme Marks
10(a)( )
9 15 8 10, 3, 1
2 2 A A
− + − = −
M1: A correct attempt to find the
midpoint between P and Q. Can be
implied by one of x or y-coordinates
correctly evaluated.M1A1
A1: ( )3, 1−
[2]
(b)
( ) ( )2 2
9 3 8 1− − + + or ( ) ( )2 2
9 3 8 1− − + +
or ( ) ( )2 2
15 3 10 1− + − + or ( ) ( )2 2
15 3 10 1− + − +
Uses Pythagoras correctly in order to find the radius. Must clearly be identified as
the radius and may be implied by their circle equation.
Or
( ) ( )2 2
15 9 10 8+ + − − or ( ) ( )2 2
15 9 10 8+ + − −
Uses Pythagoras correctly in order to find the diameter. Must clearly be identified
as the diameter and may be implied by their circle equation.
This mark can be implied by just 30 clearly seen as the diameter or 15 clearly seenas the radius (may be seen or implied in their circle equation)
Allow this mark if there is a correct statement involving the radius or the
diameter but must be seen in (b)
M1
( )( )22 2( 3) ( 1) 225 or 15 x y− + + = 2 2 2( ) ( ) x y k α β ± + ± = where ( ), A α β
and k is their radius.M1
2 2( 3) ( 1) 225 x y− + + = Allow 2 2 2( 3) ( 1) 15 x y− + + = A1
Accept correct answer only
[3]
Alternative using 2 22 2 0 x ax y by c+ + + + =
Uses ( ), A α β ± ± and 2 22 2 0 x ax y by c+ + + + =
e.g. ( ) ( )2 22 3 2 1 0 x x y y c+ − + + + = M1
Uses P or Q and 2 22 2 0 x ax y by c+ + + + =
e.g. ( ) ( )( ) ( ) ( )( )2 2
9 2 3 9 8 2 1 8 0 215c c− + − − + + + = ⇒ = − M1
2 26 2 215 0 x x y y− + + − = A1
(c)
Distance 2 215 10= −
2 2(their ) 10r = − or a correct method
for the distance e.g.
( )1 10
their their cos sin r r −
×
M1
{ }125 5 5= = 5 5 A1[2]
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Question
NumberScheme Marks
(d)
( ) 20sin30
ARQ = or
1 1090 cos15
ARQ − = −
( )( )
20 10sin or
2 their their ARQ
r r =
×
or 1 10
90 costheir
ARQr
− = −
or 1 Part( )
cos their
c ARQ r
−
=
or 1 Part( )
90 sintheir
c ARQ
r
− = −
or ( )2 2 220 15 15 2 15 15cos ARQ= + − × × 2
or
( ) ( )2
2 215 15 10 5 2 15 10 5 cos ARQ= + − × ×
A fully correct method to find , ARQ
where their 10.r >
Must be a correct statement involvingangle ARQ
M1
41.8103... ARQ = awrt 41.8 A1
[2]
Total 9
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